Question
Two arithmetic progression have the same common difference. The difference between their $100^{\text {th }}$ terms is $100$ , What is the difference between their $1000^{\text {th }}$ terms?

Answer

Let the two A.P. is be $a_1, a_2, a_3, \ldots .$. and $b_1, b_2, b 3, \ldots .$.
$a_n=a_1+(n-1) d \text { and } b_n=b_1+(n-1) d$
Since common difference of two equations is same given difference between $100^{\text {th }}$ terms is $100$
$a_{100}-b_{100}=100$
$a_1+(99) d-b_1-99 d=100$
$a_1-b_1=100 \ldots . . .(i)$
Difference between. $1000th$ terms is
$a_{1000}-b_{1000}=a_1+(1000-1) d-\left(b_1+(1000-1) d\right)$
$=a_1+999 d-b_1-999 d$
$=a_1-b_1$
$=100(\text { from }(1))$
$\therefore$ Hence difference between $1000^{\text {th }}$ terms of two A.P. is $100 .$

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